Higher Engineering Mathematics

MAXIMA, MINIMA AND SADDLE POINTS FOR FUNCTIONS OF TWO VARIABLES 363

G

z

x

y

Figure 36.11

Substituting in equation (2) gives:

S=xy+ 2 y

(

62. 5

xy

)

+ 2 x

(

62. 5

xy

)

i.e. S=xy+

125

x

+

125

y

which is a function of two variables

∂s

∂x

=y−

125

x^2

=0 for a stationary point,

hence x^2 y= 125 (3)

∂s

∂y

=x−

125

y^2

=0 for a stationary point,

hence xy^2 = 125 (4)

Dividing equation (3) by (4) gives:

x^2 y

xy^2

=1, i.e.

x

y

=1, i.e.x=y

Substitutingy=xin equation (3) givesx^3 =125,
from which,x=5m.

Hencey=5 m also

From equation (1), (5) (5)z= 62. 5

from which, z=

62. 5

25

= 2 .5m

∂^2 S

∂x^2

=

250

x^3

,

∂^2 S

∂y^2

=

250

y^3

and

∂^2 S

∂x∂y

= 1

Whenx=y=5,

∂^2 S

∂x^2

=2,

∂^2 S

∂y^2

=2 and

∂^2 S

∂x∂y

= 1

=(1)^2 −(2)(2)=− 3

Since<0 and

∂^2 S

∂x^2

>0, then the surface areaSis

aminimum.

Hence the minimum dimensions of the container to

have a volume of 62.5 m^3 are5 m by 5 m by 2.5 m.

From equation (2),minimum surface area,S

=(5)(5)+2(5)(2.5)+2(5)(2.5)

=75 m^2

Now try the following exercise.

Exercise 145 Further problems on maxima,

minima and saddle points for functions of two

variables

1. The functionz=x^2 +y^2 +xy+ 4 x− 4 y+ 3
   has one stationary value. Determine its co-
   ordinates and its nature.

[Minimum at (−4, 4)]

2. An open rectangular container is to have a
   volume of 32 m^3. Determine the dimensions
   and the total surface area such that the total
   surface area is a minimum.
   [
   4mby4mby2m,
   surface area=48m^2

]

3. Determine the stationary values of the
   function

f(x,y)=x^4 + 4 x^2 y^2 − 2 x^2 + 2 y^2 − 1

and distinguish between them.

[

Minimum at (1, 0),

minimum at (−1, 0),

saddle point at (0, 0)

]

4. Determine the stationary points of the sur-
   facef(x,y)=x^3 − 6 x^2 −y^2.
   [
   Maximum at (0, 0),
   saddle point at (4, 0)

]